SOLUTION: Identify the vertex , axis of symmetry, and direction of opening for y=2(x+3)^2-5. Thanks

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Question 43486: Identify the vertex , axis of symmetry, and direction of opening for
y=2(x+3)^2-5.
Thanks
Found 2 solutions by Nate, venugopalramana:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+2%28x+%2B+3%29%5E2+-+5
This is a vertical parabola. Since the a value is positive, the parabola opens upward.
Vertex: (h,k) for form: y+=+a%28x+-+h%29%5E2+%2B+k
Vertex: (-3,-5)
For the axis of symmetry, look at the x-value in the vertex.
Axis of Symmetry: x+=+-3
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+2x%5E2+%2B+12x+%2B+13%29+

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Identify the vertex , axis of symmetry, and direction of opening for
y=2(x+3)^2-5.
axis of symmetry is x+3=0...or....x=-3...then y=-5 .hence vertex is (-3,-5)
y =-5 for x=-3 is the minimum value .hence direction of opening is upwards for the curve.